Seq Function Erroring Out w/ Multiple Rows in R-CodePudding

I'm trying to generate rows of sequences, within a function, for each row in my dataset. The problem I'm running into is that the seq function errors out because I have multiple rows. I've included an example dataset and my code (sequence call is third from bottom of my function).

df<-data.frame(
  'Acct'=c("A","B","A"),
  'Rate'=c(8,8,12),
  'Amount'=c(1000,1000,1500),
  'Freq'=c(2,2,2),
  'MtM'=c(6,6,12),
  'YtM2'=c(.10,.10,.05),
  'periods'=c(12,12,24),
  'Price'=c(911.54,911.54,1050.37),
  'Date'=c('Sep 2021','Sep 2021', 'May 2021')
)

 dur <- function(Rate, periods,YtM2, Price ,MtM,Amount) {
  i <- 1:periods
  cf <- c(rep(Rate, periods - 1), Amount   Rate)
  pv <- (cf / (1   YtM2) ^ i)
  weight<-pv/Price
  seqi<-seq(MtM/periods,MtM,length.out=periods)
  endResults<-sum(seqi*weight)
  return(seqi) 
  }
  
  dur(df$Rate,df$periods,df$YtM2,df$Price,df$MtM,df$Amount)

When I run the code I get an error:

Error in seq.default(MtM/periods, MtM, length.out = periods) : 
  'from' must be of length 1

Ideally, I would just store these sequences temporarily. I'm trying to experiment by using dplyr::group_by statement where I group by Acct and Date and then generate the sequence ie seqi<-df%>%group_by(Acct,Date)%>%seq(MtM/df$periods,df$MtM, length.out=df$periods) but I just get this error

Error in seq.default(., MtM/df$periods, df$MtM, length.out = df$periods) : 'from' must be of length 1 In addition: Warning message: In seq.default(., MtM/df$periods, df$MtM, length.out = df$periods) : first element used of 'length.out' argument

CodePudding user response：

If you want to apply dur function for each row, you can use any of the apply function.

For example, with Map.

with(df, Map(dur, Rate, periods,YtM2, Price ,MtM,Amount))

#[[1]]
# [1] 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

#[[2]]
# [1] 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

#[[3]]
# [1]  0.5  1.0  1.5  2.0  2.5  3.0  3.5  4.0  4.5  5.0  5.5  6.0  6.5
#[14]  7.0  7.5  8.0  8.5  9.0  9.5 10.0 10.5 11.0 11.5 12.0

I removed the last Price argument from the function since it was repeated twice.

CodePudding user response：

We may use do.call with Map after subsetting the columns with the order of the input arguments in 'dur'

 do.call(Map, c(f = dur, unname(df[c(2, 7, 6, 8, 5, 3)])))
[[1]]
 [1] 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

[[2]]
 [1] 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

[[3]]
 [1]  0.5  1.0  1.5  2.0  2.5  3.0  3.5  4.0  4.5  5.0  5.5  6.0  6.5  7.0  7.5  8.0  8.5  9.0  9.5 10.0 10.5 11.0 11.5 12.0